Lecture 6 ECEN 4517/5517 Experiment 4: inverter system 12 VDC Battery HVDC: 120 - 200 VDC DC-DC converter DC-AC inverter + Isolated flyback H-bridge vac(t) AC load 120 Vrms 60 Hz – d(t) Feedback controller d(t) Vref –+ Step-up dc-dc converter with isolation (flyback) DC-AC inverter (H-bridge) Feedback controller to regulate HVDC ECEN 4517 Digital controller 1 Due dates Right now: Prelab assignment for Exp. 3 Part 3 (one from every student) Due within five minutes of beginning of lecture This week in lab (Feb. 19-21): Nothing due. Try to finish Exp. 3. Next week in lecture (Feb. 26): Prelab assignment for Exp. 4 Part 1 (one from every student) Next week in lab (Feb. 26-28): Definitely finish Exp. 3, and begin Exp. 4 The following week in lab (Mar. 4-6): Exp. 3 final report due ECEN 4517 2 Goals in upcoming weeks Exp. 4: A three-week experiment vHVDC Exp. 4 Part 1: Design and fabrication of flyback transformer Snubber circuit Demonstrate flyback converter power stage operating open loop Vbatt snubber PWM Compensator Exp. 4 Part 2: Design feedback loop Measure loop gain, compare with simulation and theory Demonstrate closed-loop control of converter output voltage ECEN 4517 3 –+ Vref Exp. 4, Part 3 H-bridge inverter, off grid vHVDC IR3101 IR3101 iac(t) + • Filtering of ac output not explicitly shown AC load 120 Vrms 60 Hz vac(t) – Digital controller • IR 3101 half-bridge modules with integrated drivers Exp. 4 Part 3: off-grid inverter • Grid-tied: control iac(t) • Demonstrate modified sine-wave inverter (required) • Off-grid: control vac(t) • Demonstrate PWM inverter (extra credit) ECEN 4517 4 “Modified Sine-Wave” Inverter vac(t) vac(t) has a rectangular waveform Inverter transistors switch at 60 Hz, T = 8.33 msec + VHVDC DT/2 T/2 – VHVDC • Choose VHVDC larger than desired Vac,RMS RMS value of vac(t) is: Vac,RMS = 1 T T 2 vac t dt = D V HVDC 0 • Can regulate value of Vac,RMS by variation of D • Waveform is highly nonsinusoidal, with significant harmonics ECEN 4517 5 PWM Inverter Average vac(t) has a sinusoidal waveform Inverter transistors switch at frequency substantially higher than 60 Hz vac(t) t • Choose VHVDC larger than desired Vac,peak • Can regulate waveshape and value of Vac,RMS by variation of d(t) • Can achieve sinusoidal waveform, with negligible harmonics • Higher switching frequency leads to more switching loss and need to filter high-frequency switching harmonics and commonmode currents ECEN 4517 6 The buck-boost converter 1 Vg 2 iL + – + Switch in position 1: Vg charges inductor V Switch in position 2: energy stored in inductor is transferred to output – Conversion ratio: V =– D Vg 1–D Subinterval 1 iL Vg + – Subinterval 2 + iL Vg V V – – ECEN 4517 + – + 7 See also: supplementary notes on Flyback converter, Exp. 4 web page The flyback converter: A transformer-isolated buck-boost converter Q1 buck-boost converter: D1 – Vg + – L V + Q1 construct inductor winding using two parallel wires: D1 1:1 Vg + – L – V + ECEN 4517 8 Derivation of flyback converter, cont. Isolate inductor windings: the flyback converter Q1 D1 – 1:1 Vg + – V LM + Flyback converter having a 1:n turns ratio and positive output: 1:n Vg + – LM Q1 ECEN 4517 9 + D1 C V – A simple transformer model Multiple winding transformer i1(t) n1 : n2 Equivalent circuit model i1(t) i2(t) + + + v1(t) v2(t) v1(t) – – i1'(t) n1 : n 2 + iM(t) LM v2(t) – – i3(t) i3(t) + v3(t) + v1(t) v2(t) v3(t) n 1 = n 2 = n 3 = ... 0 = n 1i 1' (t) + n 2i 2(t) + n 3i 3(t) + ... v3(t) – – : n3 : n3 ECEN 4517 i2(t) Ideal transformer 10 The magnetizing inductance LM Transformer core B-H characteristic • Models magnetization of transformer core material B(t) ∝ • Appears effectively in parallel with windings • If all secondary windings are disconnected, then primary winding behaves as an inductor, equal to the magnetizing inductance saturation slope ∝ LM H(t) ∝ i M (t) • At dc: magnetizing inductance tends to short-circuit. Transformers cannot pass dc voltages • Transformer saturates when magnetizing current iM is too large ECEN 4517 v1(t) dt 11 Volt-second balance in LM The magnetizing inductance is a real inductor, obeying i1(t) di (t) v1(t) = L M M dt + v1(t) integrate: i M (t) – i M (0) = 1 LM t 0 ECEN 4517 Ts 0 n1 : n 2 i2(t) + iM(t) LM v2(t) – v1(τ)dτ Magnetizing current is determined by integral of the applied winding voltage. The magnetizing current and the winding currents are independent quantities. Volt-second balance applies: in steady-state, iM(Ts) = iM(0), and hence 0= 1 Ts i1'(t) – i3(t) + v3(t) – : n3 Ideal transformer v1(t)dt 12 The “flyback transformer” Transformer model ig + – Vg i + LM vL 1:n + iC D1 C R v – – Q1 A two-winding inductor Symbol is same as transformer, but function differs significantly from ideal transformer Energy is stored in magnetizing inductance Magnetizing inductance is relatively small Current does not simultaneously flow in primary and secondary windings Instantaneous winding voltages follow turns ratio Instantaneous (and rms) winding currents do not follow turns ratio Model as (small) magnetizing inductance in parallel with ideal transformer ECEN 4517 13 Subinterval 1 Transformer model ig Vg + – i + LM vL + iC 1:n C R v – – vL = V g iC = – v R ig = i CCM: small ripple approximation leads to Q1 on, D1 off ECEN 4517 vL = V g iC = – V R ig = I 14 Subinterval 2 Transformer model ig =0 Vg + – + vL – i i/n – v/n + + iC 1:n C R v – vL = – nv i C = ni – v R ig = 0 CCM: small ripple approximation leads to vL = – V n i C = nI – V R ig = 0 Q1 off, D1 on ECEN 4517 15 CCM Flyback waveforms and solution vL Vg Volt-second balance: vL = D Vg + D' – V n =0 –V/n Conversion ratio is M(D) = V = n D Vg D' Charge balance: I/n – V/R iC i C = D – V + D' nI – V = 0 R R –V/R Dc component of magnetizing current is ig I I = nV D'R Dc component of source current is 0 DTs Conducting devices: ECEN 4517 Ts Q1 D'Ts I g = i g = D I + D' 0 t D1 16 Equivalent circuit model: CCM Flyback vL = D Vg + D' – V n =0 i C = D – V + D' nI – V = 0 R R Vg + – + I Ig DI + DV g – D'V n D'I n + – I g = i g = D I + D' 0 D' : n + I Ig + – R V – ECEN 4517 V – 1:D Vg R 17 Step-up DC-DC flyback converter Need to step up the 12 V battery voltage to HVDC (120-200 V) How much power can you get using the parts in your kit? Key limitations: MOSFET on-resistance (90 m) Input capacitor rms current rating: 25 V 2200 μF: 2.88 A 35 V 2200 μF: 3.45 A Snubber loss Need to choose turns ratio, as well as D, fs, to minimize peak currents Possible project for expo: build a better (and more complex) step-up dc-dc converter ECEN 4517 18 Design of CCM flyback transformer iM(t) n1 : n2 i1 Vg + – iM + LM vM + D1 C R V 0 i1(t) IM – – i2 Q1 0 i2(t) n1 I n2 M 0 vM(t) 0 ECEN 4517 ΔiM IM Transformer model 19 Vg DTs Approach Use your PQ 32/20 core Choose turns ratio n2/n1, LM, D, and fs (choose your own values, dont use values in supplementary notes) Select primary turns n1 so that total loss Ptot in flyback transformer is minimized: Ptot = Pfe + Pcu = core loss plus copper loss Determine air gap length Determine primary and secondary wire gauges Make sure that core does not saturate ECEN 4517 20 Core loss CCM flyback example B-H loop for this application: The relevant waveforms: B(t) B(t) Bsat Bmax Bmax ΔB 0 Hc(t) vM(t) Minor B–H loop, CCM flyback example 0 ΔB Vg n1 A c Vg DTs B–H loop, large excitation B(t) vs. applied voltage, from Faradays law: dB(t) vM (t) = n1 A c dt Solve for B: ECEN 4517 21 For the first subinterval: VgDTs B= 2n 1A c Vg dB(t) = n1 A c dt Calculation of ac flux density and core loss Fitting an equation to the plot at right P fe = K fe(ΔB) β A c l m = slope Kfe = constant that depends on fs Aclm = core volume At 60˚C: = 2.6 Kfe = 16 (50 kHz), 40 (100 kHz) with Pfe in watts, Aclm in cm3, B in Tesla From previous slide: VgDTs B= 2n 1A c More turns less B less core loss ECEN 4517 22 Copper loss Power loss in resistance of wire Must allocate the core window area between the various windings Winding 1 allocation α1WA Winding 2 allocation α2WA 0 < αj < 1 α1 + α2 + { { Total window area WA etc. Optimum choice: αm = n mI m ∞ Σ nI n=1 (leads to minimum total copper loss) j j ρ(MLT )n 21 I 2tot The resulting total copper loss is: Pcu = WAK u Choose wire gauges: ECEN 4517 + αk = 1 α 1 K uW A n1 α K W A w2 ≤ 2 u A n2 A w1 ≤ k with I tot = Σ j=1 nj n1 I j More turns more resistance more copper loss 23 Total power loss Ptot = Pcu + Pfe Power loss Co ss P c r lo fe ss P u P fe = K fe(ΔB) β A c l m re l o ppe Ptot = Pfe + Pcu ρ(MLT )n 21 I 2tot Pcu = WAK u VgDTs B= 2n 1A c Ptot Co There is a value of B (or n1) that minimizes the total power loss Optimum ΔB Prelab assignment for next week: use a spreadsheet or other computer tool to compute Ptot vs. n1, and find the optimum n1. Then design your flyback transformer. ECEN 4517 ΔB 24 Effect of transformer leakage inductance Transformer model Ll ig + vl – i 1:n LM Vg + D1 C R + – v – Q1 + – Voltage spike caused by leakage inductance di vl = L l l dt { • Ll induces a voltage spike across Q1 Vg + v/n iRon t DTs ECEN 4517 • Leakage inductance is effectively in series with transistor Q1 • When MOSFET switches off, it interrupts the current in Ll vT(t) vT(t) • Leakage inductance Ll is caused by imperfect coupling of primary and secondary windings 25 If the peak magnitude of the voltage spike exceeds the voltage rating of the MOSFET, then the MOSFET will fail. Protection of Q1 using a voltage-clamp snubber Snubber { Flyback transformer ig – Rs Vg + – vs 1:n Cs + D1 C R v + – Q1 + vT(t) • Snubber provides a place for current in leakage inductance to flow after Q1 has turned off • Peak transistor voltage is clamped to Vg + vs • vs > V/n – • Energy stored in leakage inductance (plus more) is transferred to capacitor Cs, then dissipated in Rs Usually, Cs is large Decreasing Rs decreases the peak transistor voltage but increases the snubber power loss See supplementary flyback notes for an example of estimating Cs and Rs ECEN 4517 26

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